We first make the simple observation that for non-product input distributions, the communication complexity can be arbitrarily smaller if we partition the \(k\)  inputs into \(t < k\)  players.
Indeed, consider the \(k\) -player set disjointness problem in which the \(i\) -th player, \(1 \le i \le k\) , has a set \(S_i \subseteq [n]\) ,
where for notational simplicity we define \([n]:={\left\lbrace {1,2,\dots , n} \right\rbrace }\)  for \(n\in \mathbb {N}\) .
The input distribution satisfies the promise that either (1) \(S_i \cap S_j = \emptyset \)  for every \(1\le i < j \le k\) , or (2) there is a unique item \(a \in [n]\)  such that \(a \in S_i\)  for all \(i \in [k]\) , and for any other \(a^{\prime } \ne a\) , there is at most one \(i \in [k]\)  for which \(a^{\prime } \in S_i\) . It is well-known that the randomized communication complexity of this problem is \({\Omega \left( {n/k} \right)}\)  [1]}, [2]}, [3]}, and that the bound holds even for multiple rounds of communication and players share a common blackboard.
However, if we look at \(t < k\)  players and an arbitrary, even if the worst-case mapping of the input sets \(S_1, \ldots , S_k\)  to the \(t\)  players, then by the pigeonhole principle there exists a player who gets two input sets \(S_i, S_j\)  with \(i \ne j\) .
Now this player can locally determine the output of the function by checking if \(S_i \cap S_j=\emptyset \) .
Thus with \(t < k\)  players the problem is solvable using \({{O} \left( {1} \right)}\)  bits per player. This simple argument shows that for non-product distributions, there can be an arbitrarily large gap between the \(k\) -player and the \(t\) -player worst-case-partitioned randomized communication complexities.
Note that this example applies to a symmetric problem, meaning that the \(k\) -player set disjointness problem is invariant under any one-to-one assignment of \(x_1, \ldots , x_k\)  to the \(k\)  players.
